Things are “heating up” now! Thermochemistry Things are “heating up” now!

Nature of Energy Energy – the capacity to do work (or to produce heat) Work is a force acting over a distance Moving an object Heat is a form of energy Chemicals may store potential energy in their bonds Can be released as heat energy

First Law of Thermochemistry Law of conservation of energy – energy is neither created nor destroyed Can only be converted from one form to another Potential energy – energy due to position Kinetic energy – energy due to the motion of an object KE = ½ mv2 m = mass v = velocity Units are Joules

Heat & Temperature Heat (J) Temperature (°C) Measure of energy content What is transferred during a temperature change Temperature (°C) Reflects random motion of particles in a substance Indicates the direction in which heat energy will flow Higher  lower

State Functions A property of a system that depends only on its present state Do not depend on what has happened in the system, or what might happen in the future Independent of the pathway taken to get to that state Eg: 1.0 L of water behind a dam has the same PE for work regardless of whether it flowed downhill to the dam or was taken uphill to the dam in a bucket. PE is a state function dependent only on the current position of the water

Chemical Energy Exothermic Reactions Endothermic Reactions Reactions that give off energy as they progress Some of the PE stored in the chemical bonds is converted to thermal energy (random KE) through heat Products are generally more stable (stronger bonds) than reactants Endothermic Reactions Reactions in which energy is absorbed from surroundings Energy flows into the system to increase the PE of the system Products are generally less stable (weaker bonds) than the reactants Energy is needed to break bonds; released when formed

Thermodynamics System Energy ΔE = q + w q = heat w = work Positive in endothermic reactions Negative in exothermic reactions w = work Negative if the system does work Positive if work is done on the system

Example Calculate the ΔE for a system undergoing an endothermic process in which 18.7 kJ of heat flows and where 4.3 kJ of work is done on the system. ΔE = q + w q = 18.7 kJ (endothermic – gaining heat) w = 4.3 kJ (positive – work is done on the system) ΔE = 18.7 kJ + 4.3 kJ = 23.0 kJ System has gained 23.0 kJ of energy

Thermodynamics - Gases Work done by gases w = -PΔV By a gas (through expansion) ΔV is positive w is negative To a gas (by compression) ΔV is negative w is positive P is the P of surroundings

Example Calculate the work associated with the expansion of a gas from 32 L to 75 L at a constant external pressure of 15 atm. w = -PΔV P = 15 atm ΔV = 75 L – 32 L = 43 L w = -(15 atm)(43 L) w = -645 L · atm -w because the gas expands so work is done to the surroundings Energy flows out of the system

Enthalpy & Calorimetry Enthalpy – measure of the energy released/absorbed by a substance when bonds are broken/formed during a reaction Way to calculate heat flow in or out of a system H = E + PV In systems at constant pressure, where the only work is PV, the change in enthalpy is due only to energy flow as heat (ΔH = heat of reaction) ΔH = Hproducts – Hreactants Exothermic reactions: -ΔH Endothermic reactions: +ΔH

Calorimetry Calorimetry – the science of measuring heat Heat capacity (c) Ratio of heat absorbed to increase in temperature c = heat absorbed temperature increase Specific heat capacity – energy required to raise the temperature of 1 gram of a substance by 1°C Molar heat capacity – energy required to raise the temperature of 1 mole of a substance by 1°C

q = mcΔT q = heat m = mass c = specific heat ΔT = change in T

Constant Pressure Calorimetry (soln) Calculating Heat of Reaction (ΔH) ΔH = (s)(m) (ΔT) ΔH = (specific heat capacity) x (mass of soln) x (T change) Heat of reaction is an extensive property Dependent on the amount of substance ΔH α moles of reactant

Constant Volume Calorimetry Volume of bomb calorimeter cannot change No work is done Bomb calorimetry = constant volume The heat capacity of the calorimeter must be known kJ/°C Bomb calorimetry – weighed reactants are placed inside a steel container and ignited Used by industry to determine number of food calories that we consume “Calories” on food labels are really kilocalories

ΔE = q + w, where w = 0  ΔE = q

Example – Bomb Calorimetry Heat capacity of calorimeter = 11.3 kJ/°C Compare heat of combustion for: 1.50 g methane T increase = 7.3°C 1.15 g hydrogen 14.3°C 0.5269 g octane 2.25°C Hcombustion CH4 = 54.0 kJ/g Hcombustion H2 = 140.5 kJ/g Hcombustion C8H18 = 48.3 kJ/g

Hess’s Law

Hess’s Law Hess’s Law - In going from a particular set of reactants to a particular set of products, the change in enthalpy (ΔH) is the same whether the reaction takes place in one step or a series of steps

Hess’s Law One step: Two step: N2(g) + 2 O2(g)  2 NO2(g) ΔH = 68 kJ Two step: N2(g) + O2(g) 2 NO(g) ΔH1 = 180 kJ 2 NO(g) + O2(g)  2 NO2(g) ΔH2 = -112 kJ N2(g) + 2 O2(g)  2 NO2(g) ΔH1 + ΔH2 = 68 kJ

Characteristics of Enthalpy Changes If a reaction is reversed, the sign on ΔH is reversed N2(g) + 2 O2(g)  2 NO2(g) ΔH = 68 kJ 2 NO2(g)  N2(g) + 2 O2(g) ΔH = -68 kJ Magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer

Using Hess’s Law Work backward from the final reaction Reverse reactions as needed Also reverse ΔH Identical substances found on both sides of the summed equation cancel each other

Example Given the following data Calculate ΔH for the reaction NH3(g)  ½ N2(g) + 3/2 H2(g) ΔH = 46 kJ 2 H2O(g)  2 H2(g) + O2(g) ΔH = 484 kJ Calculate ΔH for the reaction 2 N2(g) + 6 H2O(g)  3 O2(g) + 4 NH3(g)

Standard Enthalpies of Formation ΔHf°

Standard Enthalpy of Formation The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all elements in their standard state. ΔHf° - Degree sign indicates that the process was carried out under standard conditions Can measure enthalpy changes only by performing heat-flow experiments

Standard State For a compound For an element Gaseous state Pressure of 1 atm Pure liquid or solid Standard state is the pure liquid or solid Substance in solution Concentration of 1 M For an element The form in which the element exists at 1 atm and 25°C

ΔHf° always given per mole of product with the product in its standard state ½ N2(g) + O2(g)  NO2(g) ΔHf° = 34 kJ/mol C(s) + 2 H2(g) + ½ O2(g)  CH3OH(l) ΔHf° = -239 kJ/mol

Calculating Enthalpy Change Elements in their standard states are not included ΔHf° = 0 The change in enthalpy for a reaction can be calculated from the enthalpies of formation of the reactants and products ΔH°reaction = npΔH°f(products) - nrΔH°f(reactants)

Calculating ΔHf° CH4(g) + O2(g)  CO2(g) + 2 H2O(l) Using Hess’s Law – break apart the reactants into their respective elements CH4(g)  C(s) + 2 H2(g) C(s) + 2 H2(g)  CH4(g) ΔHf° = -75 kJ/mol O2(g)  O2(g) ΔHf° = 0 kJ/mol C(s) + O2(g)  CO2(g) ΔHf° = -394 kJ/mol H2(g) + ½ O2(g)  H2O(l) ΔHf° = -286 kJ/mol 2 H2O required: (-286 kJ/mol)(2) = ΔHf° = -572 kJ

Calculating ΔHf° ΔHf° = (-ΔHf° CH4(g)) + (ΔHf° O2(g)) + (ΔHf° CO2(g)) + 2(ΔHf° H2O(l)) ΔHf° = -(-75 kJ) + (0) + (-394 kJ) + (-572 kJ) ΔHf° = -891 kJ

ΔHf° If ΔHf° for a compound is negative, energy is released when the compound is formed from pure elements and the product is more stable that its constituent elements Exothermic process If ΔHf° for a compound is positive, energy is absorbed when the compound is formed from pure elements and the product is less stable than its constituent elements Endothermic process